Teaching and Learning Structural Engineering

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AC 2009-167: TEACHING AND LEARNING STRUCTURAL ENGINEERINGANALYSIS WITH MATLAB

ansgar neuenhofer, California State Polytechnic University, San Luis Obispo

© American Society for Engineering Education, 2009



Teaching and Learning Structural Engineering Analysiswith MATLAB

Abstract

The paper presents several examples of how the author uses the analysis and visualization

software MATLAB in teaching analysis courses in a structural engineering university program.

It also addresses how students work with the software to accelerate learning and deepen

understanding.

Examples include:

• Animating the vibration behavior of a building frame during earthquake excitation with

and without base isolation (only conference presentation, not included in paper).

•

Assembling structural stiffness matrices in the direct stiffness and finite element methods. • Visualizing stresses in a beam with holes.

• Behavior of a shear wall with openings under lateral loading.

• Illustrating the idea behind influence lines using a truss bridge.

It is shown how the many built-in matrix and graphics functions enable students to focus on the

logic behind an analysis method rather than time-consuming programming and debugging. The

paper includes a remarkably effortless and efficient technique in MATLAB for finite element

mesh generation of areas with openings. Shape and number of openings are arbitrary provided

we can describe them by mathematical functions.

1.0 Introduction

MATLAB1 is a software tool with powerful computational and graphics presentation capabilities

widely used in education and research. It is valuable for teaching structural analysis, in particular

modern matrix procedures like the direct stiffness and finite element methods. The popularity of

MATLAB in teaching analysis in structural engineering is due to its ease of use through a variety

of built-in functions well suited for structural analysis and its powerful graphic capabilities.

The author has taught structural analysis classes at California Polytechnic State University in San

Luis Obispo (Cal Poly) for many years and has used MATLAB in his research and for his lecture

notes. His students have worked with the software for their homework assignments and senior

projects. The main objective of having the students work with MATLAB is to help them appre-ciate principles and concepts of structural analysis and the clarity of their formulation and to

teach them to think logically. MATLAB is also ideal for understanding structural behavior

through parametric studies. In special purpose commercial structural analysis software, changing

parameters often requires cumbersome clicking through numerous levels of windows. In MAT-

LAB, on the other hand, users enjoy changing input information in an "old-fashioned" text file,

the MATLAB m-file.



2.0 Basics of finite element analysis in MATLAB

In what follows we give a brief overview of key MATLAB statements that are used to carry out

important computational steps in a finite element analysis.

2.1 Adding element stiffness matrix to structure stiffness matrix

Figure 1: Sample finite element mesh and four-node element with eight degrees-of-freedom

Let us consider the simple finite element mesh above consisting of 80 four-node elements and 99

nodes. We use the standard four-node element with two degrees-of-freedom per node (eight de-

grees-of-freedom total), the displacements in the horizontal and vertical directions. The structure

thus has 198 degrees-of-freedom. For the sake of argument, we place the 8x8 element stiffness

matrixk of element 24 (the element highlighted above) into the 198x198 structure stiffness ma-trixK . Since element 24 has nodes 26, 27, 38 and 37, the proper location of the element stiffness

matrix in the structure stiffness matrix is defined by the eight-component “location” vector

2 26 1 2 26 2 27 1 2 27 2 38 1 2 38 2 37 1 2 37

51 52 53 54 75 76 73 74

⎡ ⎤= ⋅ − ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅ − ⋅⎢ ⎥⎣ ⎦

⎡ ⎤= ⎢ ⎥⎣ ⎦

loc

The location vector is the fundamental piece of information required in the assembly loop thatcalculates the structure stiffness matrixK (see following statements).

1d

1 2

34

2d 4d

3d

5d

6d 8d

7d

x L

y L

x

y

2

F 2

F F



K = sparse(dim,dim); %dim = 198for element = 1:numele %numele = 80

%calculate element stiffness matrixkele = elestiffness()

%calculate location vector

loc = ....

K(loc,loc) = K(loc,loc) + keleend

Students usually struggle for some time before they grasp the important statement above, which

is one of the key statements in a finite element analysis. Understanding begins, once the student

recognizes MATLAB’s submatrix capabilities, i.e. that the location vector loc temporarily redu-

ces the 198x198 structure stiffness matrix to a 8x8 submatrix to which the element stiffness ma-

trix is added.

Figure 2 provides an illustration of the critical statement above, i.e. how we assemble the indivi-

dual element stiffness matrices to form the structure stiffness matrix. For simplicity, each of the

six nodes of the eight-element structure is assumed to have only one degree-of-freedom. Eachquartet of x-symbols thus represents the 2x2 element stiffness matrix. An empty box stands for a

zero entry in the structure stiffness matrix. We observe that a matrix element ij a is non-zero only

if the corresponding nodes i and j are directly connected by an element. Figure 2 schematically

depicts the state of the 6x6 structure stiffness matrix each time one of the eight elements, re-

presented by their 2x2 element stiffness matrix, is added to it. Several x-symbols in one box

indicate that the corresponding numerical values should be added.

When the structure stiffness matrix assembly process is animated in class according to Fig. 2,

students generally grasp the underlying concept quickly. As mentioned before, it requires much

more effort to convert this visual understanding into appropriate programming statements as the

ones above.



Figure 2: Illustration of structure stiffness matrix assembly

Element 1 Element 2

Element 3 Element 4

Element 5 Element 6

Element 7 Element 8



2.2 Displacement boundary conditions

To efficiently incorporate the displacement boundary conditions into MATLAB (we consider

only homogeneous boundary conditions here), we use the find command to locate the fixed

degrees-of-freedom (see Fig. 1).

%boundary condition at pin connection

fixnode = find(x==0 & y==0)

fixdof = [2*fixnode-1 2*fixnode];

%boundary condition at roller connection

fixnode = find(x==Lx & y==0)

fixdof = [fixdof 2*fixnode];

In order to eliminate the fixed degrees-of-freedom from the stiffness matrix it is practical to

specify the complementary set, the free degrees-of-freedom, which are obtained from the fixed

degrees-of-freedom by the operations

dof = zeros(1,dim) %dim= number of dofs(fixed and free combined)

dof(fixdof)=1;

free = find(dof==0);

We reduce the structure force vector F and the structure stiffness matrixK to form the correspon-

ding quantities freeF and FreeK , solve the set of linear equations for the vector freeq of structure dis-

placements and finally add the prescribed zero displacements to the solution vector using the

following statements.

%initialize displacement vector

q = zeros(dim,1);

%reduce stiffness matrix (eliminate rows and columns representing fixed dofs)

Kfree = K(free,free);

%reduce force vector

Ffree = F(free);

%solve equations

qfree = Kfree \ Ffree;

%include fixed degrees of freedom in displacement vector

q(free) = qfree;



Similarly to the statement

K(loc,loc) = K(loc,loc) + kele

discussed before, which adds the element stiffness matrix to the structure stiffness matrix,

students need time to understand the sequence of commands above involving sub-matrix and

sub-vector operations.

2.3 Applying forces to finite element model

We apply the load to the model (see Fig. 1) by using the find command again.

%distributed load in vertical direction (2nd degree of freedom at loaded

%nodes)

loadnode = find(y==Ly); %find all nodes on top of beam

F(2*loadnode) = F;

%load at corner points

loadnode = find(y==Ly & (x==0 | x==Lx));

F(2*loadnode) = F/2;



3.0 Example applications

3.1 General remarks

In what follows we present two examples of student assignments in a first course on finite ele-

ment analysis for structural engineers in the Architectural Engineering Master Program at CalPoly San Luis Obispo. The instructor provides students with a core version of a MATLAB pro-

gram containing functions that calculate the x -andy - coordinates of a rectangular grid, the con-

nectivity information, i.e. the four nodes defining each element, the element stiffness matrix, the

displacement-to-strain transformation matrix and the elasticity matrix. Students then perform

parametric studies varying the finite element mesh density and other structural parameters.

This section also contains an illustration generated in MATLAB to visualize the concept of in-

fluence lines that the author uses in a classical third-year structural analysis class.

3.2 Beam with holes

Figure 3: (a) Beam with holes. (b) Corresponding finite element model

In the first example students investigate how the number of holes and their diameter relative to

the depth of the cross section affect the stiffness and the maximum stresses in the beam. The

plane stress four-node element with incompatible modes is used. The interested reader finds a

MATLAB function calculating the element stiffness matrix in the Appendix. Stiffness and maxi-

mum stress of the perforated beam are compared to the corresponding values for a beam without

holes obtained from conventional beam theory. Students experiment with a brute force method ofmeshing around the holes by starting with a finite element mesh of the solid beam and then cal-

culating the distance from the center of each circle to the midpoint of all elements. Only those

elements are retained in the analysis whose midpoint lies outside of all circles. If the finite ele-

ment mesh is fine, this extraordinary simple and efficient procedure, which constitutes a particu-

lar highlight of this project, approximates the circles well (see Fig. 3). A summary of the corres-

ponding MATLAB statements is provided below.

H D

w

/3L /3L /6L/6L

(a)

(b)



%CALCULATE COORDINATES OF CENTER OF EACH CIRCLE delcir = Lx/numcir; %distance between centers of circles for cir = 1:numcir;

XR(cir) = delcir/2 + (cir-1)*delcir;YR(cir) = Ly/2;

end

%CALCULATE DISTANCE FROM CENTER OF EACH CIRCLE TO MIDPOINT OF ALL %ELEMENTS AND COMPARE TO RADIUS OF CIRCLE %RETAIN ELEMENTS THAT LIE OUTSIDES OF ALL CIRCLES eleretain = [1:numele]; %initialize for i = 1:numcir

r1 = sqrt ( (xm-XR(i)).^2 + (ym-YR(i)).^2 );a1 = find( r1 > rad); %Find elements whose midpoints are outside circle i eleretain = intersect(eleretain,a1);

end

Students use MATLAB's patch and surf commands to plot the response of the beam (see

Figs. 4 and 5).

Figure 4: Deflected shape of beam with five holes ( / 0.6D H = )

Figure 5: Qualitative axial stresses (a, b) and shear stresses (c, d) for beam with and withoutholes ( / 0.6D H = )

(a)

(b)

(c)

(d)



3.3 Shear wall with opening

Most of the Architectural Engineering students at Cal Poly enter the structural engineering pro-

fession in California, which is heavily focused on seismic design. Assessing the lateral stiffness

of shear walls, an important parameter in predicting the response of shear wall buildings during

earthquakes, is thus a typical problem facing our graduates in professional practice.

The behavior of a solid prismatic wall is well understood and a sufficiently accurate structural

model is that of a deep beam. The structural behavior of shear walls containing openings, how-

ever, is significantly more complex because the openings cause disturbed regions in which

“loads turn corners” resulting in irregular stress flow such that standard deep beam theory does

not apply. The structural problem of shear walls with openings is that of general plane stress for

which analytical solutions generally do not exist (see Fig. 6).

Figure 6: (a) Qualitative deflected shape and (b) qualitative shear stresses for wall with and

without opening under lateral loading

This example considers a shear wall of variable aspect ratio /H L with a horizontally and ver-

tically centered rectangular opening of dimensions /3x /3L H as shown in Fig. 7. We choose

3.60mH = for the height of the wall and vary its length L corresponding to a selected aspect

ratio /H L . We determine the lateral stiffness of the wall using both a popular method for hand

calculation (see reference 2) and finite element analysis. Selected values for the aspect ratio

are /H L = 0.4, 0.6, 0.8,..., 3.

(a)

(b)



Figure 7: Shear wall with variable aspect ratio /H L

Figure 8(a) shows the stiffness of the wall with opening relative to that of the solid wall for both

hand calculation and finite element analysis. In the hand method, the opening reduces the stiff-

ness fairly uniformly by about 10-15% for all aspect ratios. The reduction in stiffness due to the

opening is thus not sensitive to the aspect ratio of the wall. Results of finite element analysis con-trast sharply. For the shear-dominated wall with small aspect ratio the opening reduces the stiff-

ness by almost 50% whereas for the flexure-dominated wall with large aspect ratio the stiffness

is only about 20% smaller than that of the solid wall. This suggests that the stiffness reduction

strongly depends on the aspect ratio, a remarkable finding considering that the area of the ope-

ning is( ) ( )1/3 1/3 11%⋅ = of the area of the wall regardless of the aspect ratio.

Figure 8(b) summarizes the discrepancy between hand and finite element analyses. For walls go-

verned by shear deformation, the hand procedure overestimates the lateral stiffness by as much

as 60%. Hand calculation results improve with increasing aspect ratio. For tall walls dominated

by flexural deformation the error reduces to about 15%.

H / 3

H / 3

H / 3

L /3 L /3 L /3



Figure 8: (a) Stiffness relative to that of solid wall; (b) Error of hand method

3.4 Illustration of influence lines

Influence lines provide important information in structural engineering regarding the controlling

load position of moving loads and are particularly valuable in the analysis and design of bridges.

It is important for students to distinguish between the concept of an internal force diagram and

that of an influence line. An internal force diagram shows the magnitude of an internal force, say

the axial force, in all members of the structure due to a stationary loading. An influence line, on

the other hand, is a diagram showing the variation of an internal force in a single member of the

structure as a load moves across the structure. The author uses his structural analysis program inMATLAB to visualize this concept. Each of the four diagrams in Fig. 9 constitutes an internal

force diagram. An influence line by contrast is the variation (the change in color in Fig. 9) of a

force in one single member, say the vertical bar in the leftmost triangular panel of the truss

bridge, as the load changes its position.

0 0.5 1 1.5 2 2.5 30.5

0.6

0.7

0.8

0.9

Aspect Ratio H/L

R a t i o o f S t i f f n e s s e s / 0.6H L =

/ 3.0H L =

Hand Method

FEM

0 0.5 1 1.5 2 2.5 310

20

30

40

50

60

Aspect Ratio H/L

E r r o r [ % ]

(a)

(b)

/ 1.0H L =



Figure 9: Illustration of influence line and internal force diagram

Compression Tension0



4.0 Summary

The author considers MATLAB an invaluable tool for teaching and learning the analytical as-

pects of structural engineering. Its superb graphic capabilities and vast array of built-in functions

are particularly suited for courses on matrix structural analysis like the direct stiffness and finite

element methods. Students are relieved from often time consuming and frustrating programmingdetails and often report that implementing a structural analysis methodology in MATLAB has

been enjoyable and has helped them deepen their understanding. This perception is particularly

prevalent among Master students.

5.0 Acknowledgement

The author would like to express his sincere gratitude to his colleagues Kevin Dong and Allen

Estes, Associate Professor and Department Head, respectively, in the Architectural Engineering

Department at Cal Poly. Their tireless efforts in developing an ARCE Master Program and

obtaining its approval have given the author the opportunity to teach his much-loved finite

elements.

Bibliography

1. MATLAB, The MathWorks, Inc., Natick, MA

2. Neuenhofer A. (2006). Lateral stiffness of shear walls with openings. Journal of Structural Engineering , ASCE,

132(11), 1846-1851.



Appendix

%CALCULATES THE DISPLACEMENT-TO-STRAIN MATRIX B %FOR A 4-node RECTANGULAR PLANE STRESS ELEMENT

%INPUT:%xi, eta: element coordinates

%a, b: element length and width%incom=0: conventional formulation%incom=1: formulation with incompatible modes

%OUTPUT:%B: 3x8 matrix function [B] = BMatrixMembrane(xi,eta,a,b,incom);

if nargin==4;incom=0;end

if incom==0;B = zeros(3,8);

elseB = zeros(3,12);

end

c1 = (1 - eta)/ a;

c2 = (1 + eta)/ a;c3 = (1 - xi) / b;c4 = (1 + xi) / b;

c5 = - 4*2*xi / a^2;c6 = - 4*2*eta / b^2;

if incom==0;B(1,:) = [ -c1 0 c1 0 c2 0 -c2 0];B(2,:) = [ 0 -c3 0 -c4 0 c4 0 c3];B(3,:) = [ -c3 -c1 -c4 c1 c4 c2 c3 -c2];

elseB(1,:) = [ -c1 0 c1 0 c2 0 -c2 0 c5 0 0 0];B(2,:) = [ 0 -c3 0 -c4 0 c4 0 c3 0 0 0 c6];B(3,:) = [ -c3 -c1 -c4 c1 c4 c2 c3 -c2 0 c6 c5 0];

endB = B / 4;

%************************************%CALCULATES THE 3x3 ELASTICITY MATRIX%FOR PLANE STRESS%************************************function C = PlaneStressElastMatrix(E,nu);%INPUT:%E: Young's modulus%nu: Poisson's ratio

%OUTPUT:%C: 3x3 elasticity matrix

C = zeros(3,3);C(1,1) = 1;C(2,2) = 1;C(1,2) = nu;C(2,1) = nu;

C(3,3) = (1 - nu) ./2;C = C .* E ./ (1 - nu.^2);



%********************************************************** %8x8 ELEMENT STIFFNESS MATRIX of 4-NODE PLANE SRESS ELEMENT%**********************************************************function KK = EleStiffMembrane4Node(a,b,E,nu,thick,incom);%INPUT%a: half of element length in x-direction%b: half of element length in y-direction%E: modulus of elasticity

%nu: POISSON's ratio%thick: thickness of element (out-of-plane dimension)%incom=0: conventional formulation%incom=1: formulation with incompatible modes

%OUTPUT%KK = 8x8 ESM

if nargin==5;incom=0;end

%2 GAUSS pointsg(1) = -1 / sqrt(3);w(1) = 1;g(2) = -g(1);w(2) = 1;

%Elasticity matrix

C = PlaneStressElastMatrix(E,nu);if incom==0;KK = zeros(8,8);

else;KK = zeros(12,12);

end

%GAUSS integration ngauss = 2;for i = 1 : ngauss

xi = g(i);for j = 1 : ngauss

eta = g(j);B = BMatrixMembrane(xi,eta,a,b,incom);KK = KK + B' * C * B * w(i) .* w(j);

endend

KK = KK * thick * a * b;

%Static condensation to 8x8 matrix for incompatible modesif incom==1

dof1 = [1:8];dof2 = [9:12];K11 = KK(dof1,dof1);K12 = KK(dof1,dof2);K21 = KK(dof2,dof1);K22 = KK(dof2,dof2);KK = K11 - K12*inv(K22)*K21;

end



%******************************************************%Main program to analyze beam with holes in Section 3.2%******************************************************

%length of beamLx = 12;%depth of beamLy = 1.5;

x = [0 Lx Lx 0 ]';y = [0 0 Ly Ly ]';

rad = 0.3*Ly;numcir = 5;

%icom=1 = include extra displacement shapes in element formulation%icom=0 = exclude extra displacement shapes in element formulationincom=1;

%number of elements in x-direction nx = 120;

%number of elements in y-directionny = 30;

numele = nx * ny;

delx = Lx/nx;dely = Ly/ny;

a = delx/2;b = dely/2;

%generate topology of structure (x and y coordinates and connectivity)[x,y,nodes] = gen_grid_membrane(Lx,Ly,nx,ny);

%calculate miudpoint of elements for circle cut-out for i = 1:numele

xm(i) =sum(x(nodes(i,:))) / 4;ym(i) =sum(y(nodes(i,:))) / 4;

end

%CALCULATE COORDINATES OF CENTER OF EACH CIRCLE%distance between centers of circlesdelcir = Lx/numcir;for cir = 1:numcir;

XR(cir) = delcir/2 + (cir-1)*delcir;YR(cir) = Ly/2;

end

%CALCULATE DISTANCE FROM CENTER OF EACH CIRCLE TO MIDPOINT OF ALL%ELEMENTS AND COMPARE TO RADIUS OF CIRCLE%RETAIN ELEMENTS THAT LIE OUTSIDES OF ALL CIRCLES elein = [1:numele];for i=1:numcir

r1 = sqrt ( (xm-XR(i)).^2 + (ym-YR(i)).^2 );a1 = find( r1 > rad);elein = intersect(elein,a1);

end

nodes = nodes(elein,:);[numele aux] = size(nodes);numno = (nx+1) * (ny+1);

dim = 2*numno;E = 1600;nu = 0.2;thick = 1;



K = sparse(dim,dim);for i=1:dim;K(i,i) = eps; end F = zeros(dim,1);q = zeros(dim,1);

%distributed loadload = find( abs(y-Ly) < 0.00001);F(2*load) = -delx;

%distributed load corner pointsload = find( (abs(x) < 0.00001 & abs(y-Ly) < 0.00001 ) | (abs(x-Lx) < 0.00001 & abs(y-Ly)< 0.00001 ));F(2*load) = -delx/2;

%boundary condition in x-direction fix = find ( abs(x-Lx)< 0.0001 & abs(y < 0.0001) );fix = 2*fix;

%boundary condition in y-directionfix1 = find( (x == 0 & y==0) );fix1 = [2*fix1-1 2*fix1];fix = [fix fix1];

dof = zeros(1,dim);dof(fix) =1;free = find(dof==0);

%calculate element stiffness matrix (same for all elements)k = EleStiffMembrane4Node(a,b,E,nu,thick,incom);%calculate structure stiffness matrixfor i=1:numele

ino = nodes(i,1:4)';loc = zeros(8,1);loc(1:2:8) = 2*no-1;loc(2:2:8) = 2*no;K(loc,loc) = K(loc,loc) + k;

end

%reduce structure stiffness matrix to include boundary conditionsKfree = K(free,free);Ffree = F(free);

%solve equationsqfree = Kfree \ Ffree;

q(free) = qfree;

%draw deflected shapefigure;holdmag = 0.10 * Lx / max(abs(q));xmag = q(1:2:2*numno) * mag + x;ymag = q(2:2:2*numno) * mag + y;draw_patch(xmag(nodes'),ymag(nodes'))axis('equal')

%function stress = calculateStressesPlane2D(a,b,E,nu,thick,q,nodes,incom) stress = calculateStressesPlane2D(a,b,E,nu,thick,q,nodes,incom)

%CONTOUR PLOT OF STRESSES figure;holdfor i=1:numele

no = nodes(i,1:4)';

sxx = stress(i,:,1);sxx1 = [sxx(1) sxx(2);sxx(4) sxx(3)];xx = x(no(1:2));yy = y(no(2:3));surf(xx,yy,sxx1)

endset(gca,'fontsize',16)axis('equal')colorbarh = colorbar;set(h,'fontsize',20)shading interp



title('STRESSES \sigma_x_x [KSF]','fontsize',16)

%CONTOUR PLOT OF SHEAR STRESSES figure;holdfor i=1:numele

no = nodes(i,1:4)';sxx = stress(i,:,3);sxx1 = [sxx(1) sxx(2);sxx(4) sxx(3)];

xx = x(no(1:2));yy = y(no(2:3));surf(xx,yy,sxx1)

endset(gca,'fontsize',16)axis('equal')shading interpcolorbarh = colorbar;set(h,'fontsize',20)title('SHEAR STRESSES \sigma_x_y [KSF]','fontsize',16)

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Teaching and Learning Structural Engineering